\newproblem{lay:7_4_11}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 7.4.11}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Find the singular value decomposition of the matrix $A=\begin{pmatrix}-3 & 1 \\ 6 & -2 \\ 6 & -2\end{pmatrix}$
}{
   % Solution
	We compute $A^TA$
	\begin{center}
		$A^TA=\begin{pmatrix}81 & -27 \\ -27 & 9\end{pmatrix}$
	\end{center}
	Its eigenvalues and eigenvectors are
	\begin{center}
		$\lambda_1=90$, $\mathbf{v}_1=(0.9487,-0.3162)$ \\
		$\lambda_2=0$,  $\mathbf{v}_2=(0.3162,0.9487)$ \\
	\end{center}
	We now construct the matrices $V$ and $\Sigma$ as
	\begin{center}
		$V=\begin{pmatrix} \mathbf{v}_1 & \mathbf{v}_2\end{pmatrix}=\begin{pmatrix}0.9487 & 0.3162 \\ -0.3162 & 0.9487\end{pmatrix}$ \\
		$\Sigma=\begin{pmatrix} \sqrt{\lambda_1} & 0 \\ 0 & \sqrt{\lambda_2} \\ 0 & 0\end{pmatrix}=\begin{pmatrix} 9.4868 & 0 \\ 0 & 0 \\ 0 & 0 \end{pmatrix}$ \\
	\end{center}
	To construct the matrix $U$ we calculate for the non-zero singular values
	\begin{center}
		$\mathbf{u}_1=\frac{1}{\sigma_1}A\mathbf{v}_1=(-\frac{1}{3},\frac{2}{3},\frac{2}{3})$ \\
	\end{center}
	We now need to extend the set $\{\mathbf{u}_1\}$ to become a basis of $\mathbb{R}^3$. To do so, we add the vectors
	\begin{center}
		$\mathbf{u}_2=(\frac{2}{3},-\frac{1}{3},\frac{2}{3})$ \\
		$\mathbf{u}_3=(\frac{2}{3},\frac{2}{3},-\frac{1}{3})$ \\
	\end{center}
	The matrix $U$ is 
	\begin{center}
		$U=\begin{pmatrix} \mathbf{u}_1 & \mathbf{u}_2 & \mathbf{u}_3 \end{pmatrix}=
		   \begin{pmatrix}-\frac{1}{3} & \frac{2}{3} & \frac{2}{3}\\ \frac{2}{3} & -\frac{1}{3} & \frac{2}{3}\\ \frac{2}{3}& \frac{2}{3}& -\frac{1}{3}\end{pmatrix}$ \\
	\end{center}
	Finally, the SVD decomposition of $A$ is
	\begin{center}
		$A=U\Sigma V^T$ \\
		$\begin{pmatrix}-3 & 1 \\ 6 & -2 \\ 6 & -2\end{pmatrix}=
		  \begin{pmatrix}-\frac{1}{3} & \frac{2}{3} & \frac{2}{3}\\ \frac{2}{3} & -\frac{1}{3} & \frac{2}{3}\\ \frac{2}{3}& \frac{2}{3}& -\frac{1}{3}\end{pmatrix}
			\begin{pmatrix} 9.4868 & 0 \\ 0 & 0 \\ 0 & 0 \end{pmatrix}
			\begin{pmatrix}0.9487 & -0.3162 \\ 0.3162 & 0.9487\end{pmatrix}$
	\end{center}
}
\useproblem{lay:7_4_11}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}

